Tag Archives: Solution Manual for Mechanics of Materials An Integrated Learning System 4th Edition by Timothy Philpot

Mechanics of Materials: An Integrated Learning System 4th Edition by Timothy A. Philpot Complete Solution Manual

You can complete your purchase even if you do not have a PayPal account. Just click on the appropriate card type you have.

For your concerns, send an email to [email protected]

Mechanics of Materials: An Integrated Learning System 4th Edition Solution Manual by Engineering-Math.org

This is a PDF copy of the complete guide to the problems and exercises of the book Mechanics of Materials: An Integrated Learning System 4th Edition by Timothy A. Philpot. Expect the copy to be sent to your email address within 24 hours. If you have not heard from us within 24 hours, kindly send us a message to [email protected]

$49.00

Other similar materials are available

Book Covers

Mechanics of Materials Eleventh Edition by R.C. Hibbeler Complete Solution Manual

$49.00

Book Covers (1)

Mechanics of Materials 8th Edition by Beer, Johnston, DeWolf and Mazurek Complete Solution Manual

$49.00

Mechanics of Deformable Bodies

Mechanics of Materials: An Integrated Learning System 4th Edition Cover by Timothy Philpot

Mechanics of Materials: An Integrated Learning System, 4th Edition by Timothy A. Philpot

Strength of Materials Fourth Edition by Andrew Pytel and Ferdinand Singer

Strength of Materials 4th Edition by Andrew Pytel and Ferdinand Singer

Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy Philpot Problem P1.1

A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

Solution:

We are given the following values:

\begin{align*}
\text{Outside Diameter}, D &= 60\ \text{mm} \\
\text{Wall Thickness}, t & = 5\ \text{mm} \\
\text{Inside Diameter}, d & = D-2t = 60\ \text{mm}-2\left( 5\ \text{mm} \right) = 50\ \text{mm}\\
\text{Maximum Axial Stress}, \sigma & =200\ \text{MPa} = 200\ \frac{\text{N}}{\text{mm}^2}
\end{align*}

The cross-sectional area of the stainless-steel tube is

\begin{align*}
A & = \frac{\pi}{4}\left( D^2 - d^2 \right) \\
A & = \frac{\pi}{4}\left[ \left( 60\ \text{mm} \right)^2 - \left( 50\ \text{mm} \right)^2 \right] \\
A & = 863.938\ \text{mm}^2
\end{align*}

The normal stress in the tube can be expressed as

\sigma =\frac{P}{A}

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.

\begin{align*}
P_{max} & = \sigma _{max} A \\
P_{max} & = \left( 200\ \text{MPa} \right)\left( 863.938\ \text{mm}^2 \right)\\
P_{max} & = \left( 200\ \frac{\text{N}}{\text{mm}^2} \right)\left( 863.938\ \text{mm}^2 \right)\\
P_{max} & = 172788\ \text{N} \\
P_{max} & = 172.8\ \text{kN}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}